Question: A particle moves in the $xy$ -plane so that at any time $t\geq 0$ its coordinates are $x=t^3+4t$ and $y=t^4+2t^2$. What is the magnitude of the particle's acceleration vector at $t=1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $4\sqrt{31}$ (Choice B) B $6\sqrt{5}$ (Choice C) C $6$ (Choice D) D $2\sqrt{73}$
Explanation: Background When solving motion problems, it's important to remember the relationship between the position vector $(x,y)$, the velocity vector $\vec{v}(t)$, and the acceleration vector $\vec{a}(t)$ of the moving particle: $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$ $\vec{a}(t)=\dfrac{d}{dt}\vec{v}(t)=\left(\dfrac{d^2x}{dt^2},\dfrac{d^2y}{dt^2}\right)$ Setting up the math We are given that the particle's coordinates are $x=t^3+4t$ and $y=t^4+2t^2$, which means its position vector is $(t^3+4t, t^4+2t^2)$. We are asked to find the magnitude of the particle's acceleration vector at $t=1$. In other words, we need to find $||\vec{a}(1)||$. Finding $\vec{v}(t)$ $\begin{aligned} \vec{v}(t)&=\left(\dfrac{d}{dt}(t^3+4t),\dfrac{d}{dt}(t^4+2t^2)\right) \\\\ &=(3t^2+4,4t^3+4t) \end{aligned}$ Finding $\vec{a}(t)$ $\begin{aligned} \vec{a}(t)&=\dfrac{d}{dt}\vec{v}(t) \\\\ &=\left(\dfrac{d}{dt}(3t^2+4),\dfrac{d}{dt}(4t^3+4t)\right) \\\\ &=(6t,12t^2+4) \end{aligned}$ Finding $\vec{a}(1)$ $\begin{aligned} \vec{a}({1})&=(6({1}),12({1})^2+4) \\\\ &=(6,16) \end{aligned}$ Finding $||\vec{a}(1)||$ $\begin{aligned} ||\vec{a}(1)||&=||(C{6},{16})|| \\\\ &=\sqrt{(C{6})^2+({16})^2} \\\\ &=\sqrt{292} \\\\ &=2\sqrt{73} \end{aligned}$ In conclusion, the magnitude of the particle's acceleration vector at $t=1$ is $2\sqrt{73}$.